00:01
So for part a, given that the initial velocity vector, the initial equaling 8 .0 meters per second, j -hat, we can say that the acceleration vector a is giving us 4 .0 meters per second squared i -hat plus 2 .0 meters per second j -hat.
00:22
Then the position vector of the particle is equalling the initial velocity vector multiplied by t plus 1⁄2.
00:31
Times the acceleration vector times t squared.
00:35
And so this is equaling 8 .0 j hat t plus one -half multiplied by 4 .0 i -hat plus 2 .0 j hat multiplied by t squared.
00:52
So we can then say that the position vector is going to be equaling to 2 .0 t squared i -hat plus 8 .0 t plus 1 .0 t squared j hat and so therefore the time that corresponds to x equaling 29 meters we can see that solving the equation we find that 2 .0 t squared is equaling 29 meters and so t is found to be 3 .8 seconds and we're going to solve for the y coordinate at that time.
01:35
So continuing on for part a, the y coordinate is going to be equaling 8 .0 meters per second, multiplied by 3 .8 seconds, and this would be plus 1 .0 meters per second squared, multiplied by 3 .8 seconds quantity squared.
01:54
And so this is giving us 45 meters...